2010 AMC 12B Problem 11

Attempt Problem 11 of the 2010 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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11.

A palindrome between

1000

and

10{,}000

is chosen at random. What is the probability that it is divisible by

7?

\dfrac{1}{10}

\dfrac{1}{9}

\dfrac{1}{7}

\dfrac{1}{6}

\dfrac{1}{5}

A four-digit palindrome

\overline{abba}

equals

1001a+110b

1001=7\cdot11\cdot13

is divisible by

7,

but

110

is not

A four-digit palindrome has the form

\overline{abba}=1001a+110b

with

1\le a\le9

and

0\le b\le9.

Since

1001=7\cdot11\cdot13

is divisible by

7

and

110

is not, the number is divisible by

7

exactly when

b

is divisible by

7,

that is

b=0

b=7.

For each

a,

that is

2

of the

10

choices of

b,

a probability of

\dfrac{2}{10}=\dfrac15.

Thus, the correct answer is E.

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